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\title{Initial induction worksheet (with typo fix)}
\author{Dr. Holmes}
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I find that my approach is different enough from that of the book that I have to do some additional lecture before I can assign too many of the problems from the book. So I am writing out a couple of exercises for you for this round.
This worksheet is due on Friday October 6. The Wednesday lecture should give additional help on how to approach it.
\begin{enumerate}
\item Observe the following examples:
1=1
1+3 = 4
1+3+5 = 9
1+3+5+7 = 16
and so forth.
These strongly suggest that the sum of the first $n$ odd numbers, which can be written $\Sigma_{i=1}^n\,(2i-1)$, is equal to $n^2$. [can you see why this isn't $\Sigma_{i=1}^n\,(2i+1)$?]
Can you write a ``proof" of this in the style of Gauss's schoolboy proof that I did in my class example?
Set this up as a proof by mathematical induction. I'm sketching what you have to do here, but please write out your entire proof on a separate page.
What is our statement $P(n)$?
Write out the basis step $P(1)$: this shouldn't present any problems.
Now demonstrate the induction step: fix an arbitrary $k$. Assume (inductive hypothesis) that we have $P(1), P(2), \ldots, P(k-1)$ for each $k$. Then our goal is to show $P(k)$.
Notice that $P(k-1)$ means ``$1+3+5+\ldots+(2k-3) = (k-1)^2$". This is actually what you need to show $P(k)$.
Your final proof should look quite similar to my first proof in class.
\item Prove by induction that $n^2+5n$ is divisible by 6 for any natural number $n$.
I suggest using the alternative strategy:
Formulate the assertion $P(n)$ (I've already written it!)
Prove the basis step (easy)
Let $k$ be an arbitrary natural number (it does not have to be $\geq 2$). Assume $P(k)$ (the inductive hypothesis). Your goal is $P(k+1)$.
\end{enumerate}
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